Answer
a) $3x-y+2z=-8$
b) The intercepts are:
$x=\dfrac{-8}{3}$
$ y=8$
$z=-4$
Work Step by Step
a) The plane containing the point $P(0,2,-3)$ and having the normal vector $n=3i-j+2k=\lt 3,-1,2 \gt$ is expressed algebraically by the equation as follows:
$3(x-0) -1(y-2)+2(z-(-3))=0$
or, $3x-y+2+2z+6=0 \implies 3x-y+2z=-8$
b) To find the $x$-intercept, let us consider $y=z=0$
Thus, $3x=-8 \implies x=\dfrac{-8}{3}$
To find the $y$-intercept, let us consider $x=z=0$
Thus, $-y=-8 \implies y=8$
To find the $z$-intercept, let us consider $x=y=0$
Thus, $2z=-8 \implies z=-4$
